A roulette wheel has $38$ slots, of which $18$ are red, $18$ are black, and $2$ are green. In each round of the game, a ball is tossed in the spinning wheel and lands in a random slot. Suppose we watch $7$ rounds of this game, and let $R$ represent the number of rounds where the ball lands in a red slot. Which of the following would find $P(R=3)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$ (Choice B) B $\left( \dfrac{18}{38} \right)^4 \left( \dfrac{20}{38} \right)^3$ (Choice C) C ${38 \choose 7} \left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$ (Choice D) D ${7 \choose 3} \left( \dfrac{18}{38} \right)^4 \left( \dfrac{20}{38} \right)^3$ (Choice E) E ${7 \choose 3} \left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$
Explanation: Probability of $3$ successes We want the probability that there are $3$ successes (ball lands in red slot) in $7$ trials (number of rounds), so we're going to need $4$ failures (ball doesn't land in a red slot) as well. The probability of each success is ${\dfrac{18}{38}}$ and the probability of each failure is $\dfrac{20}{38}}$. It's reasonable to assume that rounds are independent, so we can multiply probabilities to find the probability of getting $3$ successes followed by $4$ failures: $\begin{aligned} P(\text{SSSFFFF})&=\left({\dfrac{18}{38}}\right)\left({\dfrac{18}{38}}\right)\left({\dfrac{18}{38}}\right)\left(\dfrac{20}{38}}\right)\left(\dfrac{20}{38}}\right)\left(\dfrac{20}{38}}\right)\left(\dfrac{20}{38}}\right) \\\\ &=\left({\dfrac{18}{38}}\right)^3\left(\dfrac{20}{38}}\right)^4 \end{aligned}$ The binomial coefficient ${n \choose k}$ SSSFFFF isn't the only arrangement that produces $3$ successes in $7$ trials. For instance, FFFFSSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=3$ successes (ball lands in red slot) in $n=7$ trials (number of rounds), so we should use the binomial coefficient ${7 \choose 3}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $\left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$ so for our final answer we multiply this probability by the number of possible arrangements: ${7 \choose 3} \left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$ The answer: ${7 \choose 3} \left( \dfrac{18}{38} \right)^3 \left( \dfrac{20}{38} \right)^4$